How to replace `eval` when assigning variables in Bash

Question

I have a function myFunction() that shall take an integer variable from outside, print it and then increment it.

This works:

myFunction:

myFunction() {
    local var="${1}"
    eval "printf '%s\n' \$$var"
    eval "$var=$(($var+1))"
}

But, I don't want to use eval in this as a user of this function might enter a wrong variable name that might then be executed. I would like to use something like printf -v as it would make the usage safer.

How can I do this?

The linked duplicate's answers are overkill for this case. Just use `myFunction() { printf '%s\n' "${!1}"; (($1++)); }` -- Indirect variable reference + the fact that `$` refs are expanded first in arithmetic expressions takes care of it. — Gordon Davisson, Dec 20 '20 at 09:25
@GordonDavisson Thanks a lot! This works just fine! Could you post it as an answer, I would like to accept it! — TimFinnegan, Dec 20 '20 at 09:41

score 0 · Answer 1 · answered Dec 20 '20 at 08:46

0

You can (only?) do this without eval if you are happy enumerating all possible variable names.

myFunction () {
  case $1 in
  (frobme)    printf '%s\n' $frobme; : $((++frobme));;
  (zapme)     printf '%s\n' $zapme;  : $((++zapme));;
  (*)         printf 'invalid name: %s\n' "$1" >&2;;
  esac
}

answered Dec 20 '20 at 08:46

Jens

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Thanks, but I don't know the variable names in advance. – TimFinnegan Dec 20 '20 at 09:41

How to replace `eval` when assigning variables in Bash

1 Answers1